Math 391 Problems — with Answers
Professor Chavel
To the student: The following problems are lifted from my files. Some errors may have crept
into the text during the copy/paste. Some problems might be repeated. Also, note the approach
to second order equations with regular singular points changes in midstream. My advice: Copy
a batch of problems, do them and then, give yourself a grade.
Question 1. Solve for the general solution: x4 dy/dx + 5x3 y = x5 .
Answer: We have
x4 y 0 + 5x3 y
x5 y 0 + 5x4 y
(x5 y)0
x5 y
y
=
=
=
=
=
x5 ,
x6 ,
x6 ,
x7 /7 + c,
x2 /7 + cx−5 .
The longer way to do it is to rewrite the equation as:
5
dy
+ y = x,
dx x
and solve ϕ0 = (5/x)ϕ for ϕ = x5 . One then multiplies the equation by x5 , and proceeds as above.
Question 2. Solve for the general solution: (x2 + ey ) dx + (xey + 2/y 4 ) dy = 0.
Answer. Reorganize the equation as:
x2 dx + 2y −4 dy + ey dx + xey dy = 0,
and realize that ey dx + xey dy is the total differential of xey . Then the solution is:
x3 /3 − (2/3)y −3 + xey = C.
The longer way is to set M = x2 + ey and N = xey + 2/y 4 and check that ∂M/∂y = ∂N/∂x. So, there
is a function F (x, y) so that Fx = M and Fy = N . From Fx = M one has
F (x, y) = x3 /3 + xey + h(y),
which implies
Fy = xey + h0 (y) = N (x, y) = xey + 2/y 4 .
Therefore h0 (y) = 2/y 4 which implies h(y) = −(2/3)y −3 . Therefore the solution to the equation is
F (x, y) = C , that is,
x3 /3 − (2/3)y −3 + xey = C.
Question 3. Solve for the general solution: y 00 − 2y 0 − 15y = 0, y(0) = 1, y 0 (0) = 0.
Answer: The associated polynomial problem is
m2 − 2m − 15 = 0
⇒
(m − 5)(m + 3) = 0
Therefore the general solution is given by
y(x) = Ae5x + Be−3x .
⇒
m = 5, −3.
Now input the initial conditions:
1 =
0 =
A+ B
5A − 3B,
and obtain: A = 3/8 and B = 5/8. So the solution is
y(x) =
3 5x 5 −3x
e + e
.
8
8
Question 4. A tank originally contains 500 gallons of fresh water. Then water containing 3
lb/gal of salt is poured into the tank at the rate of 5 gal/min, and the mixture is allowed to
leave at the same rate. Find the amount of salt in the tank at the end of 10 minutes.
Answer: Let S(t) denote the amount of salt in the tank at time t. Certainly, S(0) = 0.
Next, the amount of salt going in at any time t is 3 lb/gal at 5 gal/min, that is, 15 lb/min.
The amount of salt leaving the tank at time t is 5 times the concentration of salt at time t, that is,
5 · (S(t)/500) = S(t)/100. Therefore the differential equation is
dS
dt
dS
S
+
dt
100
t/100
dS
e
S
et/100
+
dt
100
d t/100
e
S
dt
et/100 S
S(t)
S
,
100
=
15 −
=
15,
=
15et/100 ,
=
15et/100 ,
= 1500et/100 + c,
= 1500 + ce−t/100 ,
0 = S(0) = 1500 + c,
S(t) = 1500(1 − e−t/100 ),
S(10) = 1500(1 − e−1/10 ).
Question 5. Suppose that the temperature of a cup of coffeee standing in a room obeys
Newton’s law of cooling. Suppose the room temperature is 70o F, at noon the coffee is 150o and
at 2PM the coffee is 90o F. What temperature was the coffee at 1PM?
Answer. If T (t) denotes the temperature of the coffee at time t, then the differential equation for T is:
dT
= −k{T − 70},
dt
T (0) = 150.
We are setting time t = 0 to be noon. The solution to the equation is:
T (t) = 70 + De−kt .
To calculate D set t = 0. Then 150 = 70 + D, so D = 80. Therefore
T (t) = 70 + 80e−kt .
We want to calculate T (1), but we first must know k. To find k we use T (2) = 90.
90 = 70 + 80e−2k
⇒
1
= e−2k
4
⇒
k=
ln 4
= ln 2.
2
So
T (t) = 70 + 80e−t(ln 4)/2
T (1) = 70 + 80e− ln 2 = 110.
⇒
Question 6. a tough problem A body is discovered dead at 2AM in a room heated at 70o F,
with body temperature measured to be 86o F (normal ‘live’ body temperature is 98.6o F). Later
at the morgue, in a room maintain at 40o F, the body is first measured to be at 60o F and 1 hour
later to be at 53F. Determine the time of death.
Answer: The differential equation for Newtonian cooling is given by
dT
= −k(T − To ),
dt
where T (t) is the temperature of the body at time t, and To is the ambient room temperature. Since the
k is unknown, therefore the measurements are taken in the morgue — to calculate k.
First, the solution to the differential equation is
T (t) = To + De−kt .
To calculate k we look at the measurements in the morgue. There, To = 40F, T (0) = 60F, which
implies D = 60 − 40 = 20. Since T (1) = 53F, we have
53 = 40 + 20e−k
⇒
k = ln (20/13).
So k is now determined.
Back to the scene of the death. There, T (0) = 98.6F, and To = 70F. Therefore, the constant D is
given by
D = T (0) − To = 98.6 − 70 = 28.6.
We want to establish the “time” at which the temperature of the body was taken. Then
86 = T (t) = 70 + 28.6e−t ln(20/13)
Therefore the person died
⇒
t=
ln (28.6)/16)
.
ln (20/13)
ln (28.6)/16)
ln (20/13)
hours before 2AM.
Question 7. Solve for the solution of the intial-value problem:
y 00 + 4y 0 + 20y = 0,
y 0 (0) = −1.
y(0) = 0,
Answer: The associated polynomial problem is
m2 + 4m + 20 = 0
⇒
m = −2 ± 4i.
Therefore the general solution is given by
y(x) = Ae−2x cos 4x + Be−2x sin 4x.
For the initial conditions y(0) = 0, y 0 (0) = −1 we have A = 0 and B = −1/4. So
1
y(x) = − e−2x sin 4x.
4
As an aside, suppose you wrote the general solution to the differential equation as:
y(x) = αe(−2+4i)x + βe(−2−4i)x ,
where α and β are complex. Then y(0) = 0 implies 0 = α + β, and y 0 (0) = −1 implies −1 = α(−2 +
4i) + β(−2 − 4i). Therefore, solving the two equations
0 =
−1 =
α + β,
α(−2 + 4i) + β(−2 − 4i) = −2(α + β) + 4i(α − β)
for α and β yields
α = i/8
and
β = −i/8;
Therefore
y(x) = (i/8)e−2x {cos 4x + i sin 4x} − (i/8)e−2x {cos 4x − i sin 4x} = −(1/4)e−2x sin 4x,
as above.
Question 8. Solve for the general solution: (e−x + y 4 cos x) dx + (4y 3 sin x + 2y) dy = 0.
Answer:
(e−x + y 4 cos x)dx + (4y 3 sin x + 2y)dy
e−x dx + (y 4 cos x dx + 4y 3 sin x dy) + 2y dy
−e−x + y 4 sin x + y 2
=
=
=
0,
0,
c.
Question 9. Solve for the general solution: x3 dy/dx + 4x2 y = e−x .
Answer:
x3 y 0 + 4x2 y
x4 y 0 + 4x3 y
(x4 y)0
=
=
=
x4 y
=
e−x ,
xe−x ,
xe−x ,
Z
xe−x dx.
But, by integration by parts,
Z
Z
−x
−x
xe dx = −xe +
e−x dx = −xe−x − e−x + c.
Therefore,
y=
1
{−xe−x − e−x + c}.
x4
Question 1. 0 Solve for the general solution: y 00 − 2y 0 − 48y = 0.
Answer: The associated polynomial equation is:
r2 − 2r − 48 = 0
⇒
(r − 8)(r + 6) = 0,
which implies r = 8 and r = −6. Therefore the general solution is given by
y = c1 e8t + c2 e−6t .
Question 11. Solve for the solution of the intial-value problem:
y 00 + 6y 0 + 10y = 0,
y 0 (0) = 2.
y(0) = 0,
Answer: The associated polynomial equation is given by
r2 + 6r + 10 = 0
⇒
(r + 3)2 + 1 = 0.
Therefore r = −3 ± i, which implies that the general solution is given by
y = Ae−3x cos x + Be−3x sin x.
Since y(0) = 0, we have A = 0. Therefore y = Be−3x sin x. Now use the initial condition y 0 (0) = 2.
Then
¯
2 = y 0 (0) = B {−3e−3x sin x + e−3x cos x}¯x=0 = B.
Therefore
y(x) = 2e−3x sin x.
Question 12. A tank originally contains 1000 gallons of fresh water. Then water containing
2 lb/gal of salt is poured into the tank at the rate of 4 gal/min, and the mixture is allowed to
leave at the same rate. Find the amount of salt in the tank at the end of 8 minutes.
Answer: Let S(t) denote the amount of salt in the tank at time t. Then the instantaneous change of
S with respect to t is the derivative S 0 (t). The total amount of salt per minute entering the tank is
(4)(2) = 8 gal/min. The density of the salt in the tank at time t is S(t)/1000 lb/gal, and the tank is
losing 4 gal/min. Therefore the tank is losing (4)(S(t)/1000) lb/min of salt. This implies
S 0 (t)
S 0 (t) +
1
250 S(t)
=
8−
4
1000 S(t),
= 8,
⇒
⇒
et/250 S 0 (t) + (et/250 /250)S(t)
=
8et/250 ,
⇒
(et/250 S(t))0
=
8et/250 ,
⇒
et/250 S(t)
=
8(250)et/250 + c,
⇒
S(t)) =
2000 + ce−t/250 .
To calculate c, we use the initial condition S(0) = 0 — we are assuming that at time t = 0 the water is
fresh. Then one concludes c = −2000. We therefore have
S(t) = 2000(1 − e−t/250 )
⇒
S(8) = 2000(1 − e−8/250 ).
Question 13. Solve for the general solution: (x + y) dx + (x + 2y) dy = 0.
Answer:
(x + y)dx + (x + 2y)dy
= 0,
x dx + (y dx + x dy) + 2y dy
x2
+ xy + y 2
2
= 0,
= c.
Question 14. Solve for the solution of the initial-value problem:
y 00 + 6y 0 + 10y = 0,
y(0) = 0,
y 0 (0) = 2.
Answer: The associated polynomial equation is given by
r2 + 6r + 10 = 0
(r + 3)2 + 1 = 0.
⇒
Therefore r = −3 ± i, which implies that the general solution is given by
y = Ae−3x cos x + Be−3x sin x.
Since y(0) = 0, we have A = 0. Therefore y = Be−3x sin x. Now use the initial condition y 0 (0) = 2.
Then
¯
2 = y 0 (0) = B {−3e−3x sin x + e−x cos x}¯x=0 = B.
Therefore
y(x) = 2e−x sin x.
Question 15. Solve for the general solution of: y 00 − 2y 0 − 3y = 3tet .
Answer: For the solution to the corresponding homogeneous equation, one first solves the polynomial
equation
0 = m2 − 2m − 3 = (m − 3)(m + 1).
Therefore the general solution to the corresponding homogeneous equation is
yh = Ae3t + Be−t .
For a particular solution to the inhomogeneous equation one tries the function
yp = (a + tb)et ,
and determines the constants a and b. Then
yp
yp0
=
=
aet + btet
aet + b(et + tet ) = (a + b)et + btet
yp00
=
(a + 2b)et + btet .
Substitute yp into the equation. Then one has
3tet
= yp00 − 2yp0 − 3yp
= (a + 2b)et + btet − 2{(a + b)et + btet } − 3(aet + btet )
= {a − 2a − 3a + 2b − 2b}et + {b − 2b − 3b}tet
= −4aet − 4btet ,
which implies
a=0
Therefore the general solution is:
and
b = −3/4.
3
y = − tet + Ae3t + Be−t .
4
Question 16. Solve for the general solution of:
3
x2 y 00 + 3xy 0 + y = x3
4
where two linearly independent solutions to the corresponding homogeneous problem are given
by
y1 (x) = x−1/2
y2 (x) = x−3/2 .
Answer: There are two ways to solve the problem. The first is to look at the equation and notice that a
particular solution of the form yp = ax3 will satisfy the equation! One easily finds that a = 4/63. which
implies that the general solution to the equation is:
y=
4 3
x + Ax−1/2 + Bx−3/2 .
63
The second method is to use variation of parameters. Here, to use the method from memory, one has to
rewrite the equation as:
3
3
y 00 + y 0 + 2 y = x,
x
4x
that is, divide the equation by the coefficient x2 of y 00 . We set the general solution to be of the form:
y(x) = A(x)x−1/2 + B(x)x−3/2 ,
where A and B satisfy
A0 x−1/2
A0 (−1/2)x3/2
Then
which implies
A=
+
+
B 0 x−3/2
B 0 (−3/2)x−5/2
A0 = x5/2
B 0 = −x7/2 ,
2 7/2
x + a,
7
2
B = − x9/2 + b.
9
= 0,
= x.
So the general solution is:
½
y
=
=
=
¾
½
¾
2 7/2
2 9/2
−1/2
x +a x
+ − x + b x−3/2
7
9
½
¾
2 2
−
x3 + ax−1/2 + bx−3/2
7 9
4 3
x + ax−1/2 + bx−3/2 .
63
Question 17. Solve for the general solution of:
(1 − t)y 00 + ty 0 − y = −3t2 e−t ,
t > 0,
where two linearly independent solutions to the corresponding homogeneous problem are given
by
y1 (t) = 1 + t,
y2 (t) = et .
Answer: We use the method of variation of parameters. So we look for solutions of the form
y(t) = A(t)(1 + t) + B(t)et ,
with the property that
A0 (1 + t) + B 0 et = o.
Then
A0 (1 + t)0 + B 0 (et )0 = −3t2 e−t ,
that is, we must solve the system of equations
A0 (1 + t) + B 0 et
A0 + B 0 et
Then
which implies
B 0 et = −A0 (1 + t)
A0 = 3te−t ,
and
⇒
=
=
0,
−3t2 e−t .
A0 − A0 (1 + t) = −3t2 e−t ,
B 0 = e−t {−A0 − 3t2 e−t } = −3{t + t2 }e−2t .
Integration by parts implies
Z
−t
A = a − 3te
+3
e−t dt = a − 3te−t − 3e−t = a − 3(1 + t)e−t ,
and
B
Z
3
3
= b + {t + t2 }e−2t −
{1 + 2t}e−2t dt
2
2
Z
3
3
3
= b + {t + t2 }e−2t + {1 + 2t}e−2t −
2e−2t dt
2
4
4
Z
3
3
3
2 −2t
−2t
= b + {t + t }e
+ {1 + 2t}e
−
2e−2t dt
2
4
4
3
3
3
= b + {t + t2 }e−2t + {1 + 2t}e−2t + e−2t
2
4
½
¾4
3
3 2 −2t
= b+
+ 3t + t e
2
2
3
= b + (1 + t)2 e−2t .
2
Therefore, the general solution is given by
y
=
=
=
3
{a − 3(1 + t)e−t }(1 + t) + {b + (1 + t)2 e−2t }et
2
a(1 + t) + bet − 3(1 + t)2 e−t + (3/2)(1 + t)2 e−t
a(1 + t) + bet − (3/2)(1 + t)2 e−t .
Question 18. Solve for the general solution of:
y (4) − 5y 00 + 4y = 0.
Answer: The associated polynomial equation is:
0 = m4 − 5m2 + 4 = (m2 − 4)(m2 − 1) = (m + 2)(m − 2)(m + 1)(m − 1).
Therefore the general solution is
y = Ae−2x + Be2x + Ce−x + Dex .
Question 19. Solve for the general solution of:
y (4) + y (3) + y 0 + y = 0.
Answer: The associated polynomial problem is:
0 = m4 + m3 + m + 1 = m3 (m + 1) + m + 1 = (m3 + 1)(m + 1)
which is solved by m = −1 and the three cube roots, in the complex plane, of −1. To solve for the cube
roots, write −1 and m in polar coordinates:
m = reiθ ,
−1 = ei(π+2πk) ,
where k ranges over all the integers. Then
r3 = 1, 3θ = π + 2πk,
⇒
r = 1, θ =
π 2π
+
k.
3
3
This produces three complex numbers:
k = 0 ↔ r = 1, θ = π/3
k = 1 ↔ r = 1, θ = π
k = 2 ↔ r = 1, θ = 5π/3
√
m = 1/2 + i 3/2,
m = −1,
√
m = 1/2 − i 3/2.
⇒
⇒
⇒
In particular, m = −1 is a double root. Therefore the general solution of the equation is:
√
√
y(t) = (At + B)e−t + et/2 (C cos 3t/2 + D sin 3t/2).
Question 20. Solve for the general solution of:
y (8) − 256y = 0.
Answer. For a solution of the form y = emx we have
m8 − 256 = 0,
so
m8 = 256ei(0+2πk) ,
⇒
m = 2eiπk/4 ,
where k varies over all the integers. Thus there are 8 distinct answers:
√
√
√
√
2, −2, ±2i, 2 ± 2i, − 2 ± 2i,
which implies that the general solution is given by
y
= Ae2x + Be−2x + C cos 2x + D sin 2x
√
√
√
√
√
√
√
√
+Ee 2x cos 2x + F e 2x sin 2x + Ge− 2x cos 2x + He− 2x sin 2x.
Question 21. Solve for the solution of the intial-value problem:
y 000 − 3y 00 + 2y 0 = et ,
y(0) = 0, y 0 (0) = −1, y 00 (0) = 0.
Answer: We start with the polynomial equation
0 = m3 − 3m2 + 2m = m(m2 − 3m + 2) = m(m − 2)(m − 1).
So the general solution to the corresponding homogeneous equation is
yh = A + Bet + Ce2t .
To find a particular solution to the inhomogeneous problem, we consider
yp = atet
⇒ yp 0 = a(et + tet ) yp 00 = a(2et + tet ) yp 000 = a(3et + tet ).
Then we have, substituting into the original equation,
et = (3 − 6 + 2)aet ,
which implies a = −1. Therefore the general solution is
y = −tet + A + Bet + Ce2t .
Now we use the initial conditions:
0 =
−1 =
0 =
0+A+B+C
−1 + 0A + B + 2C
−2 + 0A + B + 4C.
Then
B + 2C = 0
⇒
A−C =0
Therefore
A = 1,
B = −2,
− 2 + 2C = 0.
y = −tet + 1 − 2et + e2t .
C = 1,
Question 22. Solve using power series the initial-value problem:
y 0 + xy = 0,
y(0) = 1.
Then identify the solution in terms of known functions.
Answer: If one did not have to use power series, then one would say
y0
= −x,
y
which implies
ln y = −x2 /2 + c
Since y(0) = 1 then D = 1; so y = e−x
2
/2
⇒
y = De−x
2
/2
.
.
To do it with power series, one lets the solution be written as:
y
y0
=
=
∞
X
n=0
∞
X
an xn ,
nan xn−1 .
n=1
The equation then reads as
0=
∞
X
n=1
nan xn−1 +
∞
X
an xn+1 .
n=0
To write both series with coefficients of the powers xn , we have to rewrite the equation as:
0=
∞
X
n=0
(n + 1)an+1 xn +
∞
X
n=1
an−1 xn = a1 +
∞
X
n=1
{(n + 1)an+1 + an−1 }xn .
Then the recurrence equations are:
a1 = 0,
(n + 1)an+1 + an−1 = 0
n = 1, 2, . . . .
Since a1 = 0, all odd coefficients are zero. For even coefficients, the recurrence relation reads as:
−a2(k−1)
(−1)k a0
= k
,
2k
2 (k!)
a2k =
and a0 = y(0) = 1, by hypothesis. Therefore
y(x) =
∞
∞
X
X
2
(−1)k x2k
(−x2 /2)k
=
= e−x /2 .
k
2 (k!)
k!
k=0
k=0
Question 23. Use power series to solve:
y 0 − 2xy = 0,
y(0) = 32.
Be sure to identify the answer as a known function.
Answer: Set
y=
∞
X
an xn
y0 =
⇒
n=0
∞
X
nan xn−1 .
n=1
Then
0
=
=
∞
X
n=1
∞
X
nan xn−1 −
∞
X
2an xn+1
n=0
∞
X
(n + 1)an+1 xn −
n=0
=
a1 +
2an−1 xn
n=1
∞
X
{(n + 1)an+1 − 2an−1 } xn ,
n=1
which implies
a0 = 32
a1 = 0,
an+2 =
2an
n+2
for all n = 0, 1, 2, . . . .
Therefore,
a2k+1 = 0
for all k = 0, 1, 2, . . . .
For the even coefficients we have
a2(k+1) = a2k+2 =
which implies
a2k =
therefore
y = 32
∞
X
x2k
k=0
k!
a2k
2a2k
=
,
2(k + 1)
k+1
32
a0
=
;
k!
k!
= 32
∞
X
(x2 )k
k=0
k!
2
= 32ex .
Question 24. Solve the initial-value problem
y 00 + 2xy 0 + 2y = 0,
y(0) = −2, y 0 (0) = 0.
Namely, derive the appropriate recursion relation for expressing the solution as a power series;
then solve the recursion to obtain, if possible, a formula for all the coefficients. If you cannot
solve the recursion relation, then calculate the coefficients a0 , a1 , a2 , a3 , a4 , a5 ; if you can solve
the recursion relation, identify, if possible, the solution.
Answer. We use power series expanded around xo = 0. So
y
y0
y 00
=
=
=
∞
X
n=0
∞
X
n=1
∞
X
an xn
nan xn−1
n(n − 1)an xn−2 ,
n=2
which implies
0
=
=
∞
X
n=2
∞
X
n(n − 1)an xn−2 +
∞
X
n=1
=
∞
X
∞
X
2nan xn +
n=1
2a2 + 2a0 +
2a2 + 2a0 +
∞
X
n=1
∞
X
2an xn
n=0
(n + 2)(n + 1)an+2 xn +
n=0
=
2nan xn +
∞
X
2an xn
n=0
{(n + 2)(n + 1)an+2 + 2nan + 2an }xn
(n + 1){(n + 2)an+2 + 2an }xn .,
n=1
which implies
2an
for all n = 1, 2, 3, . . . .
n+2
The initial data are: a0 = −2 and a1 = 0, which implies, for odd coefficients, a2k+1 = 0 for all
k = 1, 2, 3, . . .. For the even coefficients we have
a2 = −a0 ,
a2 = −a0 ,
a4 = −
and
an+2 = −
2a2
a0
=+ ,
4
2
a6 = −
more generally,
a2k =
So
2a4
a0
=−
,
6
3·2
a8 = −
2a6
a0
=+
,
8
4·3·2
(−1)k
(−2).
k!
∞
∞
X
X
2
(−1)k x2k
(−x2 )k
y(x) = −2
= −2
= −2e−x .
k!
k!
k=0
k=0
Question 25. Find the general solution to
x2 y 00 + xy 0 − 4y = 0.
Answer. Try a solution of the form y = xr . Then r must satisfy
r(r − 1) + r − 4 = 0,
which implies r2 − 4 = 0
⇒
r = ±2. So the general solution is
y = Ax2 + Bx−2 .
Question 26. Solve for the general solution: x2 y 00 − 5xy 0 + 9y = 0.
Answer. This is an Euler equation, so we attempt a solution of the form
y = xr
xr {r(r − 1) − 5r + 9} = 0,
=⇒
that is,
0 = r2 − 6r + 9 = (r − 3)2 ,
which implies the two linearly independent solutions are
y = x3 ,
y = x3 ln x.
extra: To obtain the second solution, use “reduction of order”: That is, set
y = vx3 ,
y 0 = v 0 x3 + 3vx2
y 00 = v 00 x3 + 6v 0 x2 + 6vx,
which implies
0 =
=
v 00 x5 + 6v 0 x4 + 6vx3 − 5(v 0 x4 + 3vx3 ) + 9vx3
v 00 x5 + v 0 x4 ,
which implies
v 00
1
=−
v0
x
=⇒
ln v 0 = − ln x + C
Therefore,
=⇒
v0 =
D
x
=⇒
v = D ln x + E.
y = x3 v = Dx3 ln x + Ex3 .
Question 27. A spring is stretched 1 ft by a force of 10 lbs. A weight of 64 lb is hung from the
spring and is also attached to a viscuous damper that exerts a force of 8 lb when the velocity
of the mass is 2 ft/sec. If the mass is pulled down 2 ft below its equilibrium position and given
an initial downward velocity of 3 ft/sec, determine its position at any time t.
Answer: The differential equation governing the problem is:
mu00 + γu0 + ku = 0,
where m is the mass , γ is the damping constant, and k is the spring constant. The units we use are:
pounds for force, pounds/32 for mass, feet for length, and seconds for time.
The mass m = 64/32 = 2 lb.
The viscosity constant γ is given by: “Force=γ·velocity,” that is,
8 lb = γ2 ft/sec,
which implies γ = 4.
The spring constant k is given by
10 lb = k(1) ft,
which implies k = 10.
The equation, therefore, is
2u00 + 4u0 + 10u = 0.
The associated polynomial equation is
2m2 + 4m + 10 = 0 = 2(m2 + 2m + 5) = 2{(m + 1)2 + 4},
with roots given by
m = −1 ± 2i.
Therefore
u = Ae−t cos 2t + Be−t sin 2t.
To determine A and B we turn to the initial data. We are given
u(0) = 2,
u0 (0) = 3 .
Therefore
2
3
=
=
A,
¯
−Ae−t {cos 2t + 2 sin 2t} + Be−t {− sin 2t + 2 cos 2t}¯t=0 = −A + 2B
which implies
A = 2,
In sum,
B=
5
.
2
5
u(t) = 2e−t cos 2t + e−t sin 2t.
2
Question 28. Solve for the solution of the initial-value problem:
y 000 − 3y 00 + 2y 0 = et ,
y(0) = 0, y 0 (0) = −1, y 00 (0) = 0.
Answer: We start with the polynomial equation
0 = m3 − 3m2 + 2m = m(m2 − 3m + 2) = m(m − 2)(m − 1).
So the general solution to the corresponding homogeneous equation is
yh = A + Bet + Ce2t .
To find a particular solution to the inhomogeneous problem, we consider
yp = atet
⇒ yp 0 = a(et + tet ) yp 00 = a(2et + tet ) yp 000 = a(3et + tet ).
Then we have, substituting into the original equation,
et = (3 − 6 + 2)aet ,
which implies a = −1. Therefore the general solution is
y = −tet + A + Bet + Ce2t .
Now we use the initial conditions:
0 =
−1 =
0+A+B+C
−1 + 0A + B + 2C
0 =
−2 + 0A + B + 4C.
Then
B + 2C = 0
⇒
A−C =0
Therefore
A = 1,
B = −2,
− 2 + 2C = 0.
y = −tet + 1 − 2et + e2t .
C = 1,
Question 29. A mass that weighs 6 lb stretches a spring 3 in. The system is acted on by an
external force of 8 sin 4t lb. If the mass is pulled down by 1in, and then released, determine the
position of the mass at any time.
2
Answer: We work, in pounds-inches. Therefore, if g denotes the force of gravity, then g = (32)(12) in/sec .
So, to calculate the mass we have
6 = mg = m(32)(12)
⇒
m=
6
1
=
.
(32)(12)
64
To calculate the spring constant we have
6 = kL = 3k
The differential equaiton is:
⇒
k = 2.
1 00
u + 2u = 8 sin 4t,
64
or
u00 + 128u = 83 sin 4t.
The solution of the corresponding homogeneous equation is:
√
√
uh = A cos 128t + B sin 128t.
A particular solution is of the form
up = a sin 4t.
Then
16a + 128a = 83
⇒
a=
83
29
32
= 4 2 =
.
144
2 ·3
9
Therefore the general solution is:
u=
√
√
32
sin 4t + A cos 128t + B sin 128t.
9
Now use the initial data: u(0) = 1 and u0 (0) = 0. Then A = 1 and
0=
which implies
u=
√
32 · 4
+ 8 2B
9
⇒
16
B=− √ ,
9 2
√
√
16
32
sin 4t + cos 128t − √ sin 128t.
9
9 2
Question 30. A mass that weighs 8 lb stretches a spring 6 in. The system is acted on by an
external force of 8 sin 6t lb. If the mass is pulled down by 3in, and then released, determine the
position of the mass at any time.
2
Answer: We work, in pounds-feet. Therefore, if g denotes the force of gravity, then g = (32) ft/sec .
So, to calculate the mass we have
8 = mg = 32m
⇒
m=
1
8
= .
32
4
To calculate the spring constant we have
8 = kL = k/2
The differential equaiton is:
⇒
k = 16.
1 00
u + 16u = 8 sin 6t,
4
or
u00 + 64u = 32 sin 6t.
The solution of the corresponding homogeneous equation is:
uh = A cos 8t + B sin 8t.
A particular solution is of the form
up = a sin 6t.
Then
−36a + 64a = 32
⇒
a=
32
8
= .
28
7
Therefore the general solution is:
u=
8
sin 6t + A cos 8t + B sin 8t.
7
Now use the initial data: u(0) = 1/4 and u0 (0) = 0. Then A = 1/4 and B = − 76 , which implies
u=
8
1
6
sin 6t + cos 8t − sin 8t.
7
4
7
Question 31. Give the first five coefficients (that is, through the fourth-order term) of the
power series solution of:
y 00 + xy 0 + y = 0,
y(0) = 0, y 0 (0) = 1.
Answer. We have
y=
∞
X
an xn ,
y0 =
n=0
∞
X
nan xn−1 ,
y 00 =
n=1
∞
X
n(n − 1)an xn−2 ,
n=2
which implies
y 00 + xy 0 + y
∞
∞
∞
X
X
X
=
n(n − 1)an xn−2 +
nan xn +
an xn
0 =
=
n=2
∞
X
n=1
(n + 2)(n + 1)an+2 xn +
n=0
n=0
∞
X
nan xn +
n=1
= 2 · 1 · a2 + a0 +
∞
X
∞
X
an xn
n=0
{(n + 2)(n + 1)an+2 + (n + 1)an } xn ,
n=1
which implies
2a2 + a0 = 0,
(n + 2)(n + 1)an+2 + (n + 1)an = 0;
therefore
a2 = −
a0
,
2
an+2 = −
an
.
n+2
Note that y(0) = 0 implies a0 = 0, which implies a2k = 0 for all k = 0, 1, . . .. Also,
a1 = 1,
1
a3 = − ,
3
So
y =x−
Question 32.
x5 =
1
.
15
x5
x3
+
+ ... .
3
15
Solve the initial-value problem:
x2 y 00 − 3xy 0 + 8y = 0.
y(1) = 3, y 0 (1) = 0.
Answer: This is Euler’s equation, so we check for solutions of the form y = xr . Then we have the
indicial equation:
0 = r(r − 1) − 3r + 8 = r2 − 4r + 8 = (r − 2)2 + 4.
Therefore
r = 2 ± 2i,
which implies
x2±2i = x2 x±2i = x2 e±(2 ln x)i ,
y = Ax2 cos (2 ln x) + Bx2 sin (2 ln x).
For the initial conditions, we have
0 = y 0 (1) = 6 + 2B,
3 = y(1) = A,
which implies
Question 33.
y = 3x2 cos (2 ln x) − 3x2 sin (2 ln x).
(a) Find the roots to the indicial equation associated with:
µ
x2 y 00 − x −
¶
2
y = 0.
9
(b) Then consider the expansion corresponding to the higher root, and calculate the recurrence
relation for the coefficients. (c) Finally, find the first four coefficients of the expansion.
Answer: The associated Euler equation is
2
x2 y 00 + y = 0,
9
with indicial equation
2
2
0 = r(r − 1) + = r2 − r + =
9
9
µ
¶µ
¶
1
2
r−
r−
,
3
3
so (a) r = 1/3, 2/3. We now find the solution corresponding to r = 2/3:
y=
∞
X
n=0
an xn+2/3 ,
y0 =
∞
X
n=0
(n + 2/3)an xn−1+2/3 ,
y 00 =
∞
X
n=0
(n + 2/3)(n − 1/3)an xn−2+2/3 ,
which implies
∞
X
0 =
n=0
∞
X
=
n=0
∞
X
=
n=0
∞
X
=
(n + 2/3)(n − 1/3)an xn+2/3 −
∞
X
an xn+1+2/3 +
n=0
n+2/3
{(n + 2/3)(n − 1/3)an + 2/9} x
−
∞
X
∞
2X
an xn+2/3
9 n=0
an xn+1+2/3
n=0
∞
X
n(n + 1/3)an xn+2/3 −
an−1 xn+2/3
n=1
{n(n + 1/3)an − an−1 } xn+2/3 ,
n=1
which implies
an =
an−1
n(n + 1/3)
a1 =
a0
,
4/3
a2 =
a0
,
2(7/3)(4/3)
a0
,
3 · 2(10/3)(7/3)(4/3)
a4 =
a0
.
4 · 3 · 2(13/3)(10/3)(7/3)(4/3)
n = 1, 2, 3, . . . .
Therefore
a3 =
Question 34. Use Laplace transforms to solve the initial value problem:
y 0 − y = t 2 et
y(0) = 0.
Answer: We have
Ly 0 = sz,
Ly = z,
which implies
sz − z =
2
(s − 1)3
⇒
L{t2 et }(s) =
z=
Therefore
y = L−1 z =
2
,
(s − 1)3
2
2
3!
=
.
(s − 1)4
3! (s − 1)4
1 3 t
t e.
3
Question 35. Expand the function f (x) = x2 , 0 < x < 1, in a sine series.
Answer. Here we have
f (x) =
∞
X
bn sin πnx,
n=1
Z
bn
=
2
0
=
x2 sin πnx dx
(n = 1, 2, 3, . . .)
)
¯1
Z 1
−x2 cos πnx ¯¯
2
x cos πnx dx
2
¯ + πn
πn
0
0
(
)
¯1
Z 1
(−1)n+1 2
4
x sin πnx ¯¯
1
+
sin πnx dx
¯ − πn
πn
πn
πn
0
0
(
=
1
=
(−1)n+1 2
4
− 2 2
πn
π n
Z
1
sin πnx dx
0
=
¯1
¯
(−1)n+1 2
4
+ 3 3 cos πnx¯¯
πn
π n
0
=
(−1)n+1 2
4
+ 3 3 {(−1)n − 1} .
πn
π n
(If you took the answer to here you get full credit.) To take it a step further one has
b2k−1 =
So
x2 ∼
2
8
−
,
πn π 3 n3
b2k = −
1
,
πk
k = 1, 2, . . . .
¾
¸
∞ ·½
X
2
8
sin 2πkx
− 3 3 sin π(2k − 1)x −
.
πn π n
πk
k=1
Question 36. Find the inverse Laplace transform of:
½
−1
L
Answer: We have
¾
4s + 2
(t)
s2 − 6s + 10
4s + 2
4(s − 3) + 14
4s + 2
= 2
=
.
s2 − 6s + 10
s + 6s + 9 + 1
(s − 3)2 + 1
Therefore
½
L
−1
4s + 2
s2 − 6s + 10
¾
½
(t) = L
−1
4(s − 3) + 14
(s − 3)2 + 1
¾
(t) = 4e3t cos t + 14e3t sin t.
Question 37. Use Laplace transforms to solve the initial value problem:
y 0 − y = t 2 et
y(0) = 0.
Answer: We have
Ly = z,
which implies
sz − z =
Ly 0 = sz,
2
(s − 1)3
⇒
L{t2 et }(s) =
z=
Therefore
y = L−1 z =
2
,
(s − 1)3
2
2
3!
=
.
(s − 1)4
3! (s − 1)4
1 3 t
t e.
3
Question 38. Solve:
x2 y 00 + 5xy 0 + 8y = 0.
Answer: Of course, this is an Euler equation; so we try a solution of the form y = xr . Then one has
the indicial equation:
0 = r(r − 1) + 5r + 8 = r2 + 4r + 8 = (r + 2)2 + 4.
Therefore r = −2 ± 2i, which implies
y = Ae−2x cos (2 ln x) + Be−2x sin (2 ln, x)
is the general solution.
Question 39. Find the coefficients a0 through a4 of the power series expansion of the solution
to:
(1 + x2 )y 00 − 3y 0 + 4(cos x)y = 0,
y(0) = 1 y 0 (0) = 0.
Answer: Set
y=
∞
X
an xn = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + · · ·
n=0
Then a0 = 1 and a1 = 0, which implies
y
y0
y 00
= 1 + a2 x2 + a3 x3 + a4 x4 + · · · ,
= 2a2 x + 3a3 x2 + 4a4 x3 + · · · ,
= 2a2 + 6a3 x + 12a4 x2 + · · · .
Then
(1 + x2 )(2a2 + 6a3 x + 12a4 x2 + · · ·)
−3(2a2 x + 3a3 x2 + 4a4 x3 + · · ·)
+4(1 − x2 /2 + · · ·)(1 + a2 x2 + a3 x3 + a4 x4 + · · ·)
= 2a2 + 4 + x(6a3 − 6a2 ) + x2 (12a4 + 2a2 − 9a3 + 4) + · · · ,
0 =
which implies
2a2 + 4 = 0
6a3 − 6a2 = 0
12a4 + 2a2 − 9a3 + 4 = 0
So
⇒
⇒
⇒
a2 = −2,
a3 = −2,
a4 = −3/2.
y = 1 − 2x2 − 2x3 − (3/2)x4 + · · · .
Question 40. (a) Find the roots to the indicial equation (that is, the polynomial equation
belonging to the associated Euler equation) associated with:
2x2 y 00 − 3xy 0 + (x + 3)y = 0.
(b) Then consider the expansion corresponding to the higher root, and calculate the recurrence
relation for the coefficients. (c) Finally, find the first five coefficients (that is, a0 through a4 )
of the expansion. Give a general expression for the coefficient an .
Answer: The associated Euler equation is
2x2 y 00 − 3xy 0 + 3y = 0,
with indicial equation
2r(r − 1) − 3r + 3 = 0.
The two roots are: r = 1 and r = 3/2. So we give the solution associated with r = 3/2.
So we write
∞
X
y = x3/2
an xn =
n=0
∞
X
an xn+3/2 ,
n=0
which implies
0 =
∞
X
2(n + 3/2)(n + 1/2)an xn+3/2 − 3
n=0
(n + 3/2)an xn+3/2
n=0
+
∞
X
an xn+3/2+1 +
n=0
=
∞
X
∞
X
∞
X
3an xn+3/2
n=0
{2(n + 3/2)(n + 1/2) − 3(n + 3/2) + 3} an xn+3/2
n=0
+
=
=
∞
X
an−1 xn+3/2
n=1
∞ ½
X
¾
9
3
− 3n − + 3)an + an−1 xn+3/2
2
2
(2n2 + 8n +
n=1
∞
X
©
ª
(2n2 + 5n)an + an−1 xn+3/2 .
n=1
So
an = −
an−1
2n2 + 5n
for all n = 1, 2, . . . ,
which implies
1
a1 = − a0 ,
7
Question 41.
a2 =
1 1
a0 ,
18 7
a3 = −
1 1 1
a0 ,
33 18 7
a4 =
1 1 1 1
a0 .
52 33 18 7
Consider the differential equation
x2 y 00 + xy 0 − (1 + x2 )y = 0.
Determine the corresponding Euler equation, then the indicial equation and its roots. For the higher
root, determine the recurrence relation. If you can solve it for all the coefficients, do so and give the
series. If not, solve for a0 , . . . , a3 and give the partial sum.
Answer. The associated Euler equation is
x2 y 00 + xy 0 − y = 0,
whose associated indicial equation is
r(r − 1) + r − 1,
which has solution r = 1 and r = −1. For r = 1 we have
y
=
x
∞
X
n=0
y0
y 00
=
=
∞
X
n=0
∞
X
n=1
an xn =
∞
X
an xn+1
n=0
(n + 1)an xn
n(n + 1)an xn−1 ,
which implies
0
=
∞
X
n(n + 1)an xn+1 +
n=1
∞
X
−
=
an xn+1 −
∞
X
an xn+3
n=0
n(n − 1)an−1 xn +
n=2
∞
X
−
(n + 1)an xn+1
n=0
n=0
∞
X
∞
X
∞
X
nan−1 xn
n=1
an−1 xn +
n=1
∞
X
an−3 xn
n=3
(a0 − a0 )x + (2a1 + 2a1 − a1 )x2 +
∞
X
{n(n − 1)an−1 + nan−1 − an−1 + an−3 } xn
n=3
=
3a1 x2 +
∞
X
{(n + 1)(n − 1)an−1 + an−3 } xn ,
n=3
which implies a1 = 0 and
an−1
that is,
an
an−3
for all n ≥ 3,
(n + 1)(n − 1)
an−2
= −
for all n ≥ 2.
(n + 2)n
= −
In particular, a2 = −a0 /8 and a3 = −a1 /15 = 0; so
¾
½
x3
+ ··· .
y = a0 x −
8
Question 42.
Find the roots to the indicial equation associated with:
x2 y 00 − 2xy 0 + (x + 2)y = 0.
(b) Then consider the expansion corresponding to the higher root, and calculate the recurrence relation
for the coefficients. (c) Finally, find the first four coefficients of the expansion.
Answer. (a) The associated indicial equation is:
0 = r(r − 1) − 2r + 2 = r2 − 3r + 2 = (r − 2)(r − 1).
(remember: One can think of the given as a perturbation of an Euler equation. Which Euler equation?
x2 y 00 − 2xy 0 + 2y = 0. Near the origin, x = 0, the dominant term of x + 2 is 2.) So the roots are r = 1
and r = 2. (b) Set, for r = 2,
y=
∞
X
an xn+2
n=0
=⇒
y0 =
∞
X
(n + 2)an xn+1
=⇒
y 00 =
n=0
∞
X
(n + 2)(n + 1)an xn ,
n=0
which implies
0
=
=
x2 y 00 − 2xy 0 + (x + 2)y
∞
∞
∞
∞
X
X
X
X
(n + 2)(n + 1)an xn+2 − 2
(n + 2)an xn+2 +
an xn+3 + 2
an xn+2
n=0
n=0
n=0
n=0
∞
X
=
n=0
∞
X
=
n=1
∞
X
=
{(n + 2)(n + 1) − 2(n + 2) + 2} an xn+2 +
∞
X
an xn+3
n=0
[{(n + 2)(n + 1) − 2(n + 2) + 2} an + an−1 ] xn+2
[n(n + 1)an + an−1 ] xn+2 .
n=1
Therefore the recurrence relation is:
an = −
an−1
.
n(n + 1)
(c) The first four coefficients are
a0 ,
a1 = −
a0
,
2
a2 = +
so
a0
,
(2 · 3)2
a3 = −
an = (−1)n
a0
,
4 · 32 · 22
a4 = +
a0
;
5 · 42 · 32 · 22
a0
.
(n + 1)(n!)2
Question 43. Use Laplace transforms to solve the initial value problem:
y 00 + 4y + 5y = 2et ,
y(0) = 1, y 0 (0) = 0.
Answer. Let L denote the Laplace transform, and set
Ly = Y,
=⇒
Ly 0 = −1 + sY
=⇒
Ly 00 = 0 + sLy 0 = −s + s2 Y.
Now take the Laplace transform of the differential equation
2
1
= −s + s2 Y + 4(−1 + sY ) + 5Y = −s − 4 + {s2 + 4s + 5}Y,
s−1
which implies
Y =
s + 4 + 2/(s − 1)
(s + 4)(s − 1) + 2
s2 + 3s − 2
=
=
.
s2 + 4s + 5
(s − 1){(s + 2)2 + 1}
(s − 1){(s + 2)2 + 1}
One now writes Y in a partial fraction decomposition.
Y =
So
s2 + 3s − 2
A
B(s + 2) + C
=
+
.
2
(s − 1)(s + 2) + 1
s−1
(s + 2)2 + 1
s2 + 3s − 2 = A{(s + 2)2 + 1} + (s − 1){B(s + 2) + C}
(1)
for all s. For s = 1 one has in (1):
2 = 10A
=⇒
A=
1
.
5
For s = −2 one has in (1):
−4 = A − 3C =
1
− 3C
5
=⇒
C=
7
.
5
For s = 0 one has in (1):
½
¾
7
2
−2 = 5A − {2B + C} = 1 − 2B +
= − − 2B
5
5
=⇒
B=
4
.
5
Therefore,
Y (s) =
which implies
y(t) =
1/5
(4/5)(s + 2) + 7/5
+
,
s−1
(s + 2)2 + 1
1 t 4 −2t
7
e + e
cos t + e−2t sin t.
5
5
5
Question 44. (a) Expand the function f (x) = x, 0 ≤ x ≤ π, in a sine Fourier series. (b)
Solve the initial-boundary value problem for the heat equation on 0 ≤ x ≤ π:
uxx = ut ,
u = u(x, t), 0 ≤ x ≤ π, t > 0,
u(0, t) = u(π, t) = 0,
u(x, 0) = x.
Answer. For (a) We have
x
∞
X
=
bn sin nx,
0 ≤ x ≤ π,
n=1
bn
Z
2 π
x sin nx dx
π 0
¯π
½
¾
Z
2 −x cos nx ¯¯
1 π
cos
nx
dx
+
¯
π
n
n 0
0
=
=
2(−1)n−1
;
n
=
so
x=2
For (b), we therefore have
∞
X
(−1)n−1
sin nx,
n
n=1
0 ≤ x ≤ π.
∞
X
(−1)n−1 −n2 t
u(x, t) = 2
e
sin nx.
n
n=1
Question 45. Find the partial sum consisting of the first 4 terms of the power series expansion (that
is, a0 through a3 x3 ) of the solution to:
(1 + x2 )y 00 − 3y 0 + 4xy = 0,
Answer. We set
y(0) = 1 y 0 (0) = 0.
y = a0 + a1 x + a2 x2 + a3 x3 + · · · .
The initial conditions imply a0 = 1 and a1 = 0. Then
y
y0
=
=
y 00
=
1 + a2 x2 + a3 x3 + · · · ,
2a2 x + 3a3 x2 + · · · ,
2a2 + 6a3 x + · · · .
which implies
0 =
(1 + x2 )(2a2 + 6a3 x + · · ·) − 3(2a2 x + 3a3 x2 + · · ·) + 4x(1 + a2 x2 + a3 x3 + · · ·)
= 2a2 + x(6a3 − 6a2 + 4) + · · ·
which implies a2 = 0 and a3 = −2/3. Therefore
2
y = 1 − x3 + · · · .
3
Question 46. (a) Find the roots to the indicial equation (that is, the polynomial equation
belonging to the associated Euler equation) associated with:
µ
¶
1
x y + 2xy + x +
y = 0.
4
2 00
0
(b) Then consider the expansion corresponding to the higher root, and calculate the recurrence
relation for the coefficients. (c) Finally, Find the partial sum consisting of the first 4 terms
(that is, a0 + · · · + a3 x3 ) of the expansion.
Answer: The associated Euler equation is
1
x2 y 00 + 2xy 0 + y = 0,
4
with indicial equation
r(r − 1) + 2r +
1
= 0,
4
which implies r = −1/2 is a double root.
We therefore consider the solution
y
y0
y
∞
X
=
n=0
∞
X
=
n=0
∞
X
=
an xn−1/2 ,
(n − 1/2)an xn−3/2 ,
(n − 1/2)(n − 3/2)an xn−5/2 ,
n=0
which implies
0 =
∞
X
(n − 1/2)(n − 3/2)an xn−1/2 +
n=0
∞
X
+
∞
X
2(n − 1/2)an xn−1/2
n=0
(1/4)an xn−1/2 +
n=0
∞
X
an−1 xn−1/2
n=1
= (3/4 − 1 + 1/4)a0 +
∞
X
{[(n − 1/2)(n − 3/2) + 2(n − 1/2) + (1/4)]an + an−1 } xn−1/2
n=1
=
∞
X
{n2 an + an−1 }xn−1/2 ,
n=1
which implies
an = −an−1 /n2
for all n = 1, 2, . . . .
Therefore
a1 = −a0 ,
a2 = −a1 /4 = a0 /4,
a3 = −a2 /9 = −a0 /36,
which implies
y = a0 {1 − x +
x2
x3
−
+ · · ·}.
4
36
Question 47. Give the Fourier series for the 2π–periodic function f (x) given by
(
f (x) =
−π − x when − π < x < 0
.
π−x
when 0 < x < π
Then evaluate the Fourier at (b) x = 0, and at (c) x = π/2.
Answer. (a) First note that f (x) is an odd function with period 2π. So we only have to calculate the
sine terms.
f (x)
=
∞
X
bn sin nx,
n=1
bn
=
=
=
=
Z
2 π
(π − x) sin nx dx
π 0
½
¾
Z
2
cos nx ¯¯π 1 π
−(π − x)
cos nx dx
¯ −
π
n
n 0
0
¯
½
π¾
2 π
sin nx ¯¯
−
π n
n 2 ¯0
2
.
n
So
f (x) ∼
∞
X
2
sin nx.
n
n=1
(b) At x = 0, the Fourier is (obviously) equal to 0.
(c) At x = π/2, the Fourier series converges to the value of f (x) at x = π/2, that is f (π/2) = π/2.
Question 48. Extend the function f (x) = x2 , 0 < x < 1, in a cosine series.
Answer. Here we have
f (x) =
a0
=
an
=
=
=
=
∞
a0 X
+
an cos πnx,
2
n=1
Z π
2
2
x2 dx =
3
Z0 π
2
x2 cos πnx dx
(n = 1, 2, 3, . . .)
0
¯π
¾
½ 2
Z 1
x sin πnx ¯¯
2
x
sin
πnx
dx
2
−
¯
πn
πn 0
0
½
¾
Z 1
¯
1
4
x cos πnx ¯
1
cos πnx dx
¯ −
πn
πn
πn 0
0
(−1)n 4
4
cos
πn
=
.
π 2 n2
π 2 n2
So
x2 ∼
∞
1
4 X (−1)n
+ 2
cos πnx.
3 π n=1 n2
Question 49. Given the heat equation
uxx = ut ,
0 < x < 1, t > 0,
u(1, t) = 0,
, u(x, 0) = 3 sin πx − 4 sin 4πx.
with initial-boundary data
u(0, t) = 0,
Find the solution u(x, t).
Answer. Here we have, it in general,
u(x, t) =
∞
X
bn e−π
2
n2 t
sin πnx,
n=1
where (bn ) are the coefficients of the Fourier expansion of the initial data in a sine series on the interval
(0, π). But in our problem we are given, explicitly, that
b1 = 3,
Therefore
b4 = 4,
bn = 0 for all n 6= 1, 4.
2
2
u(x, t) = 3e−π t sin πx − 4e−16π t sin 4πx.
Final Exam, Spring 2005
Part I. Answer all questions.
Question 1. [13 points] Solve the initial value problem:
y 00 − 4y 0 + 4y = x2 + 12e2x ,
y(0) = 1, y 0 (0) = 0.
Answer. The corresponding homogeneous equation is:
y 00 − 4y 0 + 4y = 0,
which has the solution:
yh (x) = Axe2x + Be2x .
For a particular solution, one therefore sets
yp
yp 0
= ax2 + bx + c + dx2 e2 x,
= 2ax + b + d{2x2 e2 x + 2xe2x },
yp 00
= 2a + d{4x2 e2 x + 8xe2x + 2e2x },
which implies
x2 + 12e2x
= 2a + 4dx2 e2 x + 8dxe2x + 2de2x
−8ax − 4b − 8dx2 e2 x − 8dxe2x
+4ax2 + 4bx + 4c + 4dx2 e2x
= 4ax2 + (4b − 8a)x + 4c − 4b + 2a + 2de2x .
Therefore we have
4a = 1,
4b − 8a = 0,
4c − 4b + 2a = 0,
2d = 12,
which implies
a = 1/4,
b = 1/2,
c = 3/8,
d = 6.
The general solution is therefore
y = Axe2x + Be2x + (1/4)x2 + (1/2)x + 3/8 + 6x2 e2 x.
To solve the initial value problem (to determine A and B) we have
1
0
=
=
B + 3/8,
A + 2B + 1/2,
which implies
B = 5/8,
The solution is:
A = −7/4.
y = (−7/4)xe2x + (5/8)e2x + (1/4)x2 + (1/2)x + 3/8 + 6x2 e2 x.
Question 2. [8 points] Solve
µ
¶
µ
y
1
1
y cos (xy) +
dx + x cos (xy) + ln x + y
2x
2
e
¶
dy = 0.
Answer. Rewrite the equation as
µ
cos (xy){y dx + x dy} +
This implies
sin (xy) +
¶
y
1
dx + ln x dy + e−y dy = 0.
2x
2
y
ln x − e−y = C.
2
Question 3. [9 points] Find the general solution to
y 00 − 2y 0 + y =
ex
.
x
Answer: The solution to the corresponding homogeneous equation is
yh (x) = Axex + Bex ,
A, B =constants.
For the solution to the inhomogeneous equation, set
yh (x) = Axex + Bex ,
where A = A(x) and B = B(x) are now functions of x satisfying
A0 xex + B 0 ex = 0.
Then A and B must also satisfy
A0 (x + 1)ex + B 0 ex = ex /x.
We can factor ex from both equations, and obtain
A0 x + B 0
A0 (x + 1) + B 0
= 0,
= 1/x.
The solution of the two equations in the two unkowns A0 and B 0 is
A0 = 1/x
B 0 = −1,
which implies
A = ln x + C
B = −x + D.
Therefore the solution is
y(x) = (ln x + C)xex + (−x + D)ex = C 0 xex + Dex + xex ln x.
Question 4. [7 points] Solve
xy 0 − 2y = xy + xex .
Answer: Rewrite as
xy 0 − (2 + x)y
2+x
y0 −
y
x
=
xex ,
=
ex .
To find an integration factor φ, φ must satisfy
φ0
2+x
2
=−
= − − 1,
φ
x
x
which implies
ln φ = −x − 2 ln x = −x + ln x−2 ,
φ(x) = x−2 e−x .
The differential equation then becomes
d ¡ −2 −x ¢
x e y =
dx
x−2 e−x y
=
y
=
x−2 ,
1
+ C,
x
−xex + Cx2 ex .
−
Question 5. [13 points] For the equation
2xy 00 − y 0 + y = 0,
(a) Show x = 0 is a regular singular point.
(b) Find the indicial equation and the recurrence relation corresponding the larger root.
(c) Find the first four terms of the series solution valid near x > 0 corresponding to the
larger root.
Answer: First rewrite the differential equation as
y 00 −
1 0
1
y +
= 0;
2x
2x
then the associated Euler equation is:
y 00 −
1 0
y = 0,
2x
which implies the indicial equation is
1
r(r − 1) − r = 0,
2
1
2
r − r − r = 0,
2
3
2
r − r = 0.
2
The two roots are r = 0 and r = 3/2. So consider the solution associated with r = 3/2, that is,
y
=
y0
=
y 00
=
∞
X
n=0
∞
X
n=0
∞
X
an xn+3/2 ,
(n + 3/2)an xn+1/2 ,
(n + 3/2)(n + 1/2)an xn−1/2 .
n=0
Then
2xy 00 − y 0 + y
∞
X
=
2(n + 3/2)(n + 1/2)an xn+1/2
0 =
n=0
∞
X
−
+
n=0
∞
X
(n + 3/2)an xn+1/2
an xn+3/2
n=0
=
∞
X
{2(n + 3/2)(n + 1/2) − (n + 3/2)} an xn+1/2
n=0
∞
X
+
an xn+3/2
n=0
=
∞
X
(n + 3/2){(2n + 1) − 1}an xn+1/2
n=0
∞
X
+
an xn+3/2
n=0
=
∞
X
(n + 3/2)(2n)an xn+1/2
n=0
∞
X
+
an−1 xn+1/2
n=1
= 0a0 +
∞
X
{n(2n + 3)an + an−1 } xn+1/2 .
n=1
The recurrence relation is
n(2n + 3)an + an−1 = 0
for all n = 1, 2, . . . ,
that is
an = −
an−1
n(2n + 3)
a0
,
1·5
a2 = +
for all n = 1, 2, . . . .
We therefore obtain
a1 = −
which implies
½
y = a0 x
3/2
a0
2·1·7·5
a3 = −
a0
,
3·2·1·9·7·5
¾
x7/2
x9/2
x5/2
+
−
+ ··· .
−
1·5
2·1·7·5 3·2·1·9·7·5
Question 6. [4 points] Use separation of variables to replace the partial differential equation:
xtuxx + uxt + tux = 0,
where u is a function of x and t, by two ordinary differential equations.
Answer: Set u(x, t) = X(x)T (t). Then the equation becomes
xtX 00 T + X 0 T 0 + tX 0 T = 0,
where the primes denote differentiation with repsect to the single variable. Rewrite the equation as
xtX 00 T
xX 00
X0
=
=
−X 0 {T 0 + tT },
T 0 + tT
−
= λ,
tT
which implies
xX 00 − λX
T 0 + (1 + λ)tT
= 0,
= 0.
Question 6. [10 points] Use the Laplace transform method to solve:
y 00 + 4y = 2,
y(0) = 1, y 0 (0) = 3.
Answer: If (Ly)(s) = Y (s) is the Laplace transform of y, then
(Ly 0 )(s) = −y(0) + s(Ly)(s),
which implies
(Ly 0 )(s) =
−1 + sY (s),
(Ly 00 )(s) =
−3 + s(−1 + sY (s)) = −3 − s + s2 Y (s).
Taking the Laplace transform of both sides of the equation, we obtain
s2 Y − 3 − s + 4Y
=
(s2 + 4)Y
=
Y
=
2
,
s
2
+ s + 3,
s
2
s+3
+ 2
.
2
s(s + 4) s + 4
Set
2
s+3
+ 2
+ 4) s + 4
s(s2
A Bs + C
+ 2
s
s +4
A(s2 + 4) + s(Bs + C)
;
s(s2 + 4)
=
=
then the standard argument shows that
A=
1
,
2
1
,
2
B=
C = 3,
which implies
Y (s)
=
y(t)
=
1 1 s/2 + 3
+ 2
,
2s
s +4
1 1
3
+ cos 2t + sin 2t.
2 2
2
Part II. Answer any of THREE COMPLETE questions.
Question 8. [12 points] Find the Fourier series for
(
f (x) =
x+2
2−x
−2 ≤ x ≤ 0,
,
0 < x ≤ 2.
where f (x + 4) = f (x) for all x.
Answer: First note that f (x) is an even function. One can see that by drawing the graph of the
function, or by realizing that
f (x) = 2 − |x|,
−2 ≤ x ≤ 2.
Therefore all the coefficients for the sine terms vanish, and
∞
a0 X
nπx
f (x) =
+
an cos
,
2
2
n=1
where
a0
an
=
=
=
=
=
=
1
2
1
2
Z
Z
Z
2
f (x) dx =
−2
Z 2
2
(2 − x) dx = 2,
0
f (x) cos
−2
nπx
dx
2
2
nπx
dx
2
0
¯2
Z 2
2
nπx ¯¯
nπx
2
(2 − x) sin
sin
dx
+
nπ
2 ¯0 nπ 0
2
Z 2
2
nπx
sin
dx
nπ 0
2
¯2
4
nπx ¯¯
− 2 2 cos
n π
2 ¯
(2 − x) cos
0
=
=
=
4
− 2 2 {cos nπ − 1}
n π
4
{1 − (−1)n }
2
2
n
π
½ 8
n = odd
n2 π 2
.
0
n = even
Therefore
f (x) = 1 +
∞
8 X cos ((2k + 1)πx/2)
.
π2
(2k + 1)2
k=0
Question 9. [12 points] Find the terms of the power series through x4 of
y 00 − y 0 + xy = 0,
y(0) = 1, y 0 (0) = 2.
Answer: The expansion of the power series is
y(x) = 1 + 2x + a2 x2 + a3 x3 + a4 x4 + · · · ,
y 0 (x) = 2 + 2a2 x + 3a3 x2 + 4a4 x3 + · · · ,
y 00 (x) = 2a2 + 6a3 x + 12a4 x2 + · · · ,
which implies
0 = y 00 − y 0 + xy
= 2a2 + 6a3 x + 12a4 x2 + · · · − (2 + 2a2 x + 3a3 x2 + 4a4 x3 + · · ·)
+x(1 + 2x + a2 x2 + a3 x3 + a4 x4 + · · ·)
= 2a2 + 6a3 x + 12a4 x2 − 2 − 2a2 x − 3a3 x2 − 4a4 x3
+x + 2x2 + a2 x3 + a3 x4 + a4 x5 + · · ·
= 2a2 − 2 + (6a3 − 2a2 + 1)x + (12a4 − 3a3 + 2)x2 + · · · ;
Therefore
2a2 − 2 = 0,
6a3 − 2a2 + 1 = 0,
12a4 − 3a3 + 2 = 0,
which implies
a2 = 1,
a3 = 1/6,
The series is
y(x) = 1 + 2x + x2 +
a4 = −1/8.
x3
3x4
−
+ ··· .
6
2
Question 10. [12 points] A mass weighing two pounds stretches a spring 6 inches. The mass
is pulled down 3 inches and given an upward velocity of 1 ft/sec. Find u(t), the displacement
of the mass in feet from its eqiuilibrium position at time t seconds after release. assume that
the acceleration due to gravity is 32 ft/sec2 and that air resitance is negligible.
Answer: Do not forget to change inches to feet!!
The force on the spring satisfies Hooke’s law: F = kL; in our case that is
2 = k(1/2)
⇒ k = 4.
The force of gravity is given by: F = mg;in our case that is
2 = 32m
⇒ m = 1/16.
The initial value problem, therefore, is
1 00
u + 4u = 0,
16
u(0) = 1/4, u0 (0) = 1,
which implies
u00 + 64u = 0,
u(0) = 1/4, u0 (0) = 1,
Therefore,
u(0) = 1/4, u0 (0) = −1.
u(t) = A cos 8t + B sin 8t,
The initial conditions yield A = 1/4 and B = −1/8, which implies
u(t) =
1
1
cos 8t − sin 8t.
4
8
Question 11. [12 points] A 200 gallon tank is half full of pure water. A salt solution with a
concentration of 5 lb/gal is flowing into the tank at the rate of 4 gal/min while the rest of the
well-mixed solution is flowing out at the rate of 2 gal/min.
(a) Find S(t), the amount of salt in lbs in the tank at time t minutes.
(b) Find the concentration of salt in the tank when the tank overflows.
Answer: Let V (t) denote the amount of water in the tank at time t. Then V (0) = 100 and V 0 (t) =
4 − 2 = 2. So
V (t) = 100 + 2t,
and the tank overflows at time t = 50.
The initial condition for the salt S(t) is given by S(0) = 0. The rate of change of the salt is given by
dS
dt
=
=
=
lb
gal
S(t) lb
gal
·4
−
·2
gal min V (t) gal min
2S(t)
20 −
2t + 100
S(t)
20 −
,
t + 50
5
that is,
dS
dt
dS
S
+
dt
t + 50
dS
+S
(t + 50)
dt
d
((t + 50)S)
dt
S(t)
= 20 −
S
,
t + 50
= 20,
= 20(t + 50),
=
20(t + 50),
=
10(t + 50) +
S(0) = 0 implies C = 25, 000. So
S(t) = 10(t + 50) +
C
.
t + 50
25, 000
.
t + 50
The concentration at overflow is, therefore,
25 lb
S(50)
=
.
200
4 gal